Real Number
Previous Year Solved Questions 2024
1. LCM (850,500) is: (CBSE 2024)
(a) 850 × 50
(b) 17 × 500
(c) 17 × 5² × 2²
(d) 17 × 5³ × 2
(a) 850 × 50
(b) 17 × 500
(c) 17 × 5² × 2²
(d) 17 × 5³ × 2
View Answer
1. (b) 17 × 500
2. Prove that 6-4√5 is an irrational number, given that √5 is an irrational number. (CBSE 2024)
View Answer
2. Let us assume 6-4√5 = x is a rational number
⇒ √5 = (6-x)/4
Now RHS is rational but LHS is irrational.
Therefore our assumption is wrong
Hence 6-4√5 is irrational.
⇒ √5 = (6-x)/4
Now RHS is rational but LHS is irrational.
Therefore our assumption is wrong
Hence 6-4√5 is irrational.
3. Show that 11 × 19 × 23 + 3 × 11 is not a prime number. (CBSE 2024)
View Answer
3. 11 × 19 × 23 + 3 × 11
= 11 × (19 × 23 + 3)
⇒ The given number has more than two factors
Hence it is not a prime number.
= 11 × (19 × 23 + 3)
⇒ The given number has more than two factors
Hence it is not a prime number.
4. If two positive integers p and q can be expressed as p = 18a²b¹ and q = 20a³b², where a and b are prime numbers, then LCM (p, q) is: (CBSE 2024)
(a) 2a²b²
(b) 180a²b²
(c) 12a²b²
(d) 180a³b²
(a) 2a²b²
(b) 180a²b²
(c) 12a²b²
(d) 180a³b²
View Answer
4. (d) 180a³b²
5. Prove that 5-2√3 is an irrational number. It is given that √3 is an irrational number. (CBSE 2024)
View Answer
5. Assuming 5-2√3 to be a rational number.
Let 5-2√3 = a/b where a and b are integers & b ≠ 0
⇒ √3 = (5b-a)/(2b)
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, 5-2√3 is an irrational number.
Let 5-2√3 = a/b where a and b are integers & b ≠ 0
⇒ √3 = (5b-a)/(2b)
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, 5-2√3 is an irrational number.
6. Show that the number 5 × 11 × 17 + 3 × 11 is a composite number. (CBSE 2024)
View Answer
6. 5 × 11 × 17 + 3 × 11
= 11 × (5 × 17 + 3) = 11 × 88 or 11 × 11 × 2³
It means the number can be expressed as a product of two factors other than 1, therefore the given number is a composite number.
= 11 × (5 × 17 + 3) = 11 × 88 or 11 × 11 × 2³
It means the number can be expressed as a product of two factors other than 1, therefore the given number is a composite number.
7. In a teachers’ workshop, the number of teachers teaching French, Hindi and English are 48, 80 and 144 respectively. Find the minimum number of rooms required if in each room the same number of teachers are seated and all of them are of the same subject. (CBSE 2024)
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7. Minimum number of rooms required means there should be maximum number of teachers in a room.
We have to find HCF of 48, 80 and 144.
48 = 2⁴ × 3
80 = 2⁴ × 5
144 = 2⁴ × 3²
HCF(48,80,144) = 2⁴ = 16
Therefore, total number of rooms required = 48/16 + 80/16 + 144/16 = 17
We have to find HCF of 48, 80 and 144.
48 = 2⁴ × 3
80 = 2⁴ × 5
144 = 2⁴ × 3²
HCF(48,80,144) = 2⁴ = 16
Therefore, total number of rooms required = 48/16 + 80/16 + 144/16 = 17
8: Teaching Mathematics through activities is a powerful approach that enhances students’ understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class 5 students. She announced the number 2 in her class and asked the first student to multiply it by a prime number and then pass it to the second student. The second student also multiplied it by a prime number and passed it to the third student. In this way by multiplying by a prime number, the last student got 173250. Now, Mukta asked some questions as given below to the students: (CBSE 2024)
(A) What is the least prime number used by students?
(B) How many students are in the class?
OR
What is the highest prime number used by students?
(C) Which prime number has been used maximum times?
(A) What is the least prime number used by students?
(B) How many students are in the class?
OR
What is the highest prime number used by students?
(C) Which prime number has been used maximum times?
View Answer
Ans: 8
Prime factorization of 173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11
(A) Least prime number used by students = 3 (because 2 is announced by the teacher)
(B) Number of students = 7 (factors other than 2)
OR Highest prime number used by student = 11
(C) Prime number 5 is used maximum times i.e., 3 times.
Prime factorization of 173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11
(A) Least prime number used by students = 3 (because 2 is announced by the teacher)
(B) Number of students = 7 (factors other than 2)
OR Highest prime number used by student = 11
(C) Prime number 5 is used maximum times i.e., 3 times.
9. The smallest irrational number by which √20 should be multiplied so as to get a rational number, is: (CBSE 2024)
(a) √20
(b) √2
(c) 5
(d) √5
(a) √20
(b) √2
(c) 5
(d) √5
View Answer
Ans: (d) √5
√20 = 2√5
Multiplying by √5: 2√5 × √5 = 2 × 5 = 10 (rational)
√5 is the smallest irrational number that makes the product rational.
√20 = 2√5
Multiplying by √5: 2√5 × √5 = 2 × 5 = 10 (rational)
√5 is the smallest irrational number that makes the product rational.
10. The LCM of two prime numbers p and q (p>q) is 221. Then the value of 3p-q is: (CBSE 2024)
(a) 4
(b) 28
(c) 38
(d) 48
(a) 4
(b) 28
(c) 38
(d) 48
View Answer
Ans: (c) 38
Since p and q are prime numbers and LCM(p,q) = 221
221 = 17 × 13
p = 17, q = 13 (as p>q)
3p – q = 3×17 – 13 = 51 – 13 = 38
Since p and q are prime numbers and LCM(p,q) = 221
221 = 17 × 13
p = 17, q = 13 (as p>q)
3p – q = 3×17 – 13 = 51 – 13 = 38
11. A pair of irrational numbers whose product is a rational number is (CBSE 2024)
(a) (√16, √4)
(b) (√5, √2)
(c) (√3, √27)
(d) (√36, √2)
(a) (√16, √4)
(b) (√5, √2)
(c) (√3, √27)
(d) (√36, √2)
View Answer
Ans: (c) (√3, √27)
√3 × √27 = √81 = 9
9 is a rational number.
√3 × √27 = √81 = 9
9 is a rational number.
12. Given HCF(2520,6600)=40, LCM(2520,6600)=252×k, then the value of k is: (CBSE 2024)
(a) 1650
(b) 1600
(c) 165
(d) 1625
(a) 1650
(b) 1600
(c) 165
(d) 1625
View Answer
Ans: (a) 1650
Using HCF × LCM = Product of numbers
40 × (252 × k) = 2520 × 6600
k = (2520 × 6600)/(40 × 252)
k = 10 × 165 = 1650
Using HCF × LCM = Product of numbers
40 × (252 × k) = 2520 × 6600
k = (2520 × 6600)/(40 × 252)
k = 10 × 165 = 1650
Q13. Two alarm clocks ring their alarms at regular intervals of 20 minutes and 25 minutes respectively. If they first beep together at 12 noon, at what time will they beep again together next time? (CBSE 2024) [3 Marks]
View Answer
Ans 13.
Find LCM of 20 and 25:
20 = 2² × 5
25 = 5²
LCM = 2² × 5² = 4 × 25 = 100 minutes
100 minutes = 1 hour 40 minutes
12:00 PM + 1 hour 40 minutes = 1:40 PM
The alarm clocks will beep again together at 1:40 PM.
Find LCM of 20 and 25:
20 = 2² × 5
25 = 5²
LCM = 2² × 5² = 4 × 25 = 100 minutes
100 minutes = 1 hour 40 minutes
12:00 PM + 1 hour 40 minutes = 1:40 PM
The alarm clocks will beep again together at 1:40 PM.